- #1

mmapcpro

- 41

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I was looking at a problem from an old test and I got confused about something.

Q. Find the slope of the tangent line to the given curve at a given point:

y = x(x^2 + 3)^1/2 ; P(1,2)

S. First I found the derivative:

[(x^2 + 3)^1/2] + [(x^2)(x^2 + 3)^-1/2]

Plugged in the values for x and y at the given point:

[(4)^1/2] + [(4)^-1/2]

The choices for answers were:

a) 2 b) -5/2 c) 3/2 d) 5/2 e) none of these

Square root of 4 can be + OR - 2, no?

That would mean I can have 4 possible solutions:

(4/2) + (1/2) = 5/2

(4/2) + (-1/2) = 3/2

(-4/2) + (1/2) = -3/2

(-4/2) + (-1/2) = -5/2

Since 3 of these are listed as choices, how did I pick the right one? (I chose d)5/2 and it was marked correct)

Forgive me if this is shockingly stupid...I am trying to review all my calculus and physics to start classes up again in the winter.